Section 2: Answers

Hey, guys.
Section two was all about motion in two dimensions. This breaks down into projectile motion and circular motion.

1.


2.


3. C=13 cm. Θ = 26 degrees


4. (a) r = (5.00i + 12.0j) m.
(b) The magnitude of r = 13.0 m. Θ= 67.4 degrees.

5. t = 0.454 s. y = 3.3 ft.

6. (a) 3 s
(b) 45 m.

7. 1.8 m.


8. t, y max, v y (t). and x will all remain the same.


9.(a) v i x = 20 m/s. v i y = 15 m/s.
(b) v x = 20 m/s.
(c) horizontal velocity = 20 m/s
final velocity = 25 m/s at an angle of -37 degrees.
(d) and (e) see figure




10.Divide this problem into two one-dimensional problems:

(a) V o x = 20 m/s
(b) V o y = 15 m/s
(c) y max = 31.25 m/s
(d) t = 4 s
(e) x(4s) = 80 m
(f) v(4s) = 32 m/s
(g) Θ = -51.3 degrees

(a), (b), (f), and (g) are drawn on the figure below.



11. 17.8 m

12. v o = 45 m/s

13. (a) and (b) a = 10 m/s ^ 2
shown in figure



(c) neither v nor a is constant

(d) object is subjected to approximately 1 g

14.

(a) The period = 0.4 π s.
(b) The frequency = 2.5/ π s-1
(c) From (a) v = 2 πr/T.
The centripetal acceleration a = 4 π ^ 2 r f ^ 2 = (d)